Show that any $n+1$ vectors in a $n$ dimensional vector space forms a linearly dependent set

Question

$V$ is an $n$-dimensional vector space. Show that $n + 1$ vectors in $V$ form a linearly dependent set.

Here is how I am approaching it:

Let $\dim V = n$, which implies that $S$ is a linearly independent set of vectors such that $S = \{v_1,v_2,\ldots,v_n\}$ is the basis of $V$.

Let $W = \{w_1,\ldots,w_r\}$ be a set of linearly independent vectors in $V$

I don't know where to go from here.

Answer 1

The statement $\dim V = n$ means there exists a basis $\{v_1,\ldots,v_n\}$ for $V$. Let $v_{n+1} \in V\setminus \{v_1,\ldots,v_n\}$. Then by the definition of a basis, there exist $\alpha_1,\ldots,\alpha_n \in F$ ($F$ being the field over which $V$ is defined) such that $$ v_{n+1} = \alpha_1v_1 + \ldots + \alpha_nv_n. $$ So $\{v_1,\ldots,v_{n+1}\}$ is linearly dependent.

Answer 2

Believe I can answer my own question:

Assuming the above let Let $W_1$ = {$w_1$, $v_1$, $v_2$, . . . $v_n$}

$W_1$ spans V because S spans V.

$w_1$ is a linear combination of the vectors in S because S is a basis for V.

Then we get $w_1$ = $a_1$$v_1$ + $a_2$$v_2$ + . . . + $a_n$$v_n$ rearrange terms so that 0 = $a_1$$v_1$ + $a_2$$v_2$ + . . . + $a_n$$v_n$ - $w_1$ and $W_1$ is a linearly dependent set because at least one of the coefficients does not equal 0/